Finding the parameters for Harmonic Oscillator solutions

  • #1
gabu
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Homework Statement



Using the Schrödinger equation find the parameter [itex]\alpha[/itex] of the Harmonic Oscillator solution [itex]\Psi(x)=A x e^{-\alpha x^2}[/itex]

Homework Equations



[itex]-\frac{\hbar^2}{2m}\,\frac{\partial^2 \Psi(x)}{\partial x^2} + \frac{m \omega^2 x^2}{2}\Psi(x)=E\Psi(x)[/itex]

[itex]E=\hbar\omega(n+\frac{1}{2})[/itex]

The Attempt at a Solution



Using the Schrödinger equation we have arrive at

[itex] -2\alpha (2x^2\alpha-3)+\frac{m^2\omega^2x^2}{\hbar^2} = \frac{2m}{\hbar^2}E[/itex]

If I make x=0 I obtain

[itex] \alpha = \frac{m\omega}{2\hbar} [/itex]

using the information that the energy level of the oscillator is the same as the highest power in the solution, meaning [itex] E=3\hbar\omega/2 [/itex].

Now, my problem with this solution is the need to make x=0 to arrive at it. I know that the equation holds for every x, so it is justifiable to consider the origin. The thing is, however, that shouldn't I be able to solve the equation without this assumption? Shouldn't it be independent of x?

Thank you very much.
 
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  • #2
Hello.
gabu said:
Using the Schrödinger equation we have arrive at

[itex] -2\alpha (2x^2\alpha-3)+\frac{m^2\omega^2x^2}{\hbar^2} = \frac{2m}{\hbar^2}E[/itex]
This equation must be satisfied for all ##x##. This can be true only for one specific value of ##\alpha## and one specific value of ##E##.

Group together terms of like powers in ##x## to form a polynomial in ##x##. Use the fact that if a polynomial in ##x## is equal to zero for all values of ##x## in some finite domain, then each coefficient in the polynomial must be zero.
 
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  • #3
gabu said:
Using the Schrödinger equation find the parameter αα\alpha of the Harmonic Oscillator solution Ψ(x)=Axe−αx2Ψ(x)=Axe−αx2\Psi(x)=A x e^{-\alpha x^2}
This is actually the first excited state of a 1-D quantum harmonic oscillator having energy 3/2ħω
Where α is mω/2ħ
 
  • #4
TSny said:
Hello.

This equation must be satisfied for all ##x##. This can be true only for one specific value of ##\alpha## and one specific value of ##E##.

Group together terms of like powers in ##x## to form a polynomial in ##x##. Use the fact that if a polynomial in ##x## is equal to zero for all values of ##x## in some finite domain, then each coefficient in the polynomial must be zero.

Oh, I can see now. Thank you very much.
 

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